1. **Problem Statement:** Calculate the theoretical equilibrant force $E_{theo}$ and its angle $\theta_{theo}$ for Problem 1 with forces $F_1 = 125$ g at $10^\circ$ and $F_2 = 95$ g at $90^\circ$. 2. **Formula and Rules:** The equilibrant force balances the resultant of applied forces, so $\vec{E} = -\vec{R}$. To find $\vec{R}$, sum components: $$R_x = F_1 \cos \theta_1 + F_2 \cos \theta_2$$ $$R_y = F_1 \sin \theta_1 + F_2 \sin \theta_2$$ Then magnitude: $$R = \sqrt{R_x^2 + R_y^2}$$ Angle: $$\theta_R = \arctan\left(\frac{R_y}{R_x}\right)$$ The equilibrant magnitude equals $R$, direction $\theta_E = \theta_R + 180^\circ$ (opposite direction). 3. **Calculate Components:** $$F_{1x} = 125 \cos 10^\circ = 125 \times 0.9848 = 123.1$$ $$F_{1y} = 125 \sin 10^\circ = 125 \times 0.1736 = 21.7$$ $$F_{2x} = 95 \cos 90^\circ = 95 \times 0 = 0$$ $$F_{2y} = 95 \sin 90^\circ = 95 \times 1 = 95$$ 4. **Sum Components:** $$R_x = 123.1 + 0 = 123.1$$ $$R_y = 21.7 + 95 = 116.7$$ 5. **Calculate Resultant Magnitude:** $$R = \sqrt{123.1^2 + 116.7^2} = \sqrt{15156.6 + 13623.9} = \sqrt{28780.5} = 169.6$$ 6. **Calculate Resultant Angle:** $$\theta_R = \arctan\left(\frac{116.7}{123.1}\right) = \arctan(0.948) = 43.3^\circ$$ 7. **Equilibrant Force:** $$E_{theo} = R = 169.6 \text{ g}$$ $$\theta_{theo} = 43.3^\circ + 180^\circ = 223.3^\circ$$ 8. **Interpretation:** The equilibrant force has magnitude approximately 170 g and points at $223.3^\circ$, opposite the resultant. 9. **Diagram Description:** - Draw vector $F_1$ at $10^\circ$ with length proportional to 125 g. - Draw vector $F_2$ at $90^\circ$ with length proportional to 95 g. - The resultant vector $\vec{R}$ is the vector sum of $F_1$ and $F_2$. - The equilibrant vector $\vec{E}$ points opposite $\vec{R}$ with length 169.6 g. This completes the calculation and vector diagram for Problem 1.